Suppose we have a vector field $f(x, y) = (y^3 - 4y, x^3 - 4x)$ and a curve $C$ that is parameterized by $\alpha(t) = (t, 2t)$ for $0 < t < 1$. What is the line integral of $f$ along $C$ ? $ \int_C f \cdot d\alpha = $
Given a vector field $f$, a parameterization $\alpha$, and bounds $t_0$ and $t_1$, we can calculate the line integral as follows: $ \int_C f \cdot d\alpha = \int_{t_1}^{t_2} f(\alpha(t)) \cdot \alpha'(t) \, dt$ Here, $f(x, y) = (y^3 - 4y, x^3 - 4x)$ and $\alpha(t) = (t, 2t)$. $\begin{aligned} &f(\alpha(t)) = (8t^3 - 8t, t^3 - 4t) \\ \\ &\alpha'(t) = (1, 2) \end{aligned}$ Now we can rewrite our line integral as a single-variable integral. $ \int_C f \cdot d\alpha = \int_0^1 (8t^3 - 8t, t^3 - 4t) \cdot (1, 2) \, dt$ Let's solve the integral. $\begin{aligned} &\int_0^1 (8t^3 - 8t, t^3 - 4t) \cdot (1, 2) \, dt \\ \\ &= \int_0^1 8t^3 - 8t + 2t^3 - 8t \, dt \\ \\ &= \int_0^1 10 t^3 - 16t \, dt \\ \\ &= \left[ \dfrac{5}{2} t^4 - 8t^2 \right]_0^1 \\ \\ &= \dfrac{5}{2} - 8 \\ \\ &= \dfrac{-11}{2} \end{aligned}$ In conclusion, the line integral $ \int_C f \cdot d\alpha = \dfrac{-11}{2}$.